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POWhat is the most efficient way to link grid squares together?
 

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  1. POWhat is the most efficient way to link grid squares together?
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    <p>Assume that I want to represent a 10*10 room which has 100 floors. I can start from any floor and walk through the room, but, there are walls between some floors. </p> <p>CS speaking, I am trying to make a 10*10 grid -room- where each square represent a floor. Each floor has certain characteristics and it could be represented as a node, the whole grid, however contains list of these nodes. </p> <p>Im trying to link each square to the one that relates to it. The relation could be described as follow: -Each square has another one on top, right, up, down to it except the ones on the edges of the grid. </p> <p>For example, the first floor which is at the top left of the grid has relation only to the floor on its right side and the one under it. </p> <p>-Additionally, some floors can't be linked together assuming that there's a block between them. Floor no. one in the previous example can't be linked with the one under it because there's a wall between them. </p> <p>I used a pointer to link each node -square or floor- to its related nodes:</p> <pre><code>public class Node{ private Node right; private Node left; private Node up; private Node down; //constructor other methods } </code></pre> <p>However, this solution may take many places in the memory, assuming we have 100 nodes and each node has 4 pointers!</p> <p>I've changed this solution by assigning ID to each node, then in each node there is an int[] array where I can store the number of related nodes. </p> <p>This solution introduced another problem in the grid class! Assume that after the change, a new method at Node class will be : </p> <pre><code>public void setNeighbors(int[] neighbors ) { this.neighbors = neighbors; } </code></pre> <p>At the grid class when I want to create each node to add it to the list, I'll have to write 100 line, one for each node! </p> <pre><code>int [] n1 ={2}; grid.getEntry(1).setNeighbors(n1); int [] n2 ={1,3}; grid.getEntry(2).setChars(n2 ); . . . And so on.. </code></pre> <p>My question is, how can I resolve the problem by making as efficient and clean code as possible. </p> <p>How can I represent the <strong>static</strong> relation between the squares without having to create an array at each step or without having to write 100 line.</p> <p>I found a mathematical relation between the squares but I wasn't able to use it because some squares can't be linked to the onse next to it because of the wall between them..</p>
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      1. POWhat is the most efficient way to link grid squares together?
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    1. COI don't see what's wrong with the node class you posted. It makes traversal pretty easy. Could be hard to populate I guess. Depends if a node needs to know who its neighbours are or not. I really don't see what you think you're gaining by adding ID properties instead of using object references.
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    2. COI didn't like the 400 pointers idea. Plus, I'd still have to do 100 lines but it'd look like this `grid.getEntry(2).setNeighbors(grid.getEntry(1), grid.getEntry(3), null,null);`
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    3. COThe "400 pointers idea" is far better than creating 400 ids (which are essentially pointers in this context) and managing them yourself, which is your proposed alternative. You could keep the floor in a two dimensional array of nodes, create a clean node in each then iterate through the array assigning left/right/up/down nodes. No need for 100 lines of code.
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